Problem: Solve the equation. $\dfrac{dy}{dx}=-\dfrac{2y^2}{x^2}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\dfrac{x}{-2+Cx}$ (Choice B) B $y=-\dfrac{x}{2}+C$ (Choice C) C $y=\sqrt[3]{-\dfrac{x^3}{2}}+C$ (Choice D) D $y=\sqrt[3]{-\dfrac{x^3}{2}+C}$
Solution: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=-\dfrac{2y^2}{x^2} \\\\ \dfrac{1}{y^2}\,dy&=-\dfrac{2}{x^2}\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} \dfrac{1}{y^2}\,dy&=-\dfrac{2}{x^2}\,dx \\\\ \int \dfrac{1}{y^2}\,dy&=\int -\dfrac{2}{x^2}\,dx \\\\ -\dfrac{1}{y}&=\dfrac{2}{x}+C_1 \\\\ \dfrac{1}{y}&=\dfrac{-2}{x}+C \\\\ \dfrac{1}{y}&=\dfrac{-2+Cx}{x} \\\\ y&=\dfrac{x}{-2+Cx} \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\dfrac{x}{-2+Cx}$